Certainly! Here’s a rewritten version of your steps, conveying the same meaning but with different wording and formatting: --- **Step 1: Analyze the Circuit** - The circuit consists of an ideal op-amp, with two resistors: \( R_1 \) and \( R_2 \). - The input voltage is \( v_i \), and the output voltage is \( v_o \). - The non-inverting (+) terminal is connected to ground, so its voltage is 0 V. - Due to the virtual short property of an ideal op-amp, the inverting (–) terminal is also at 0 V. - Let’s refer to the inverting terminal as node A. **Step 2: Define Currents** - The current flowing through \( R_1 \) is: \[ I_1 = \frac{v_i - V_A}{R_1} = \frac{v_i - 0}{R_1} = \frac{v_i}{R_1} \] - The current through \( R_2 \) is: \[ I_2 = \frac{V_A - v_o}{R_2} = \frac{0 - v_o}{R_2} = -\frac{v_o}{R_2} \] - Since the ideal op-amp draws no current at its inputs, the current through \( R_1 \) must equal the current through \( R_2 \). **Step 3: Apply Kirchhoff’s Current Law (KCL) at Node A** - According to KCL, the sum of currents entering the node is zero: \[ I_1 + I_2 = 0 \] - Substitute the expressions for \( I_1 \) and \( I_2 \): \[ \frac{v_i}{R_1} - \frac{v_o}{R_2} = 0 \] **Step 4: Derive the Gain (\( v_o/v_i \))** - Rearranging the equation gives: \[ \frac{v_o}{R_2} = \frac{v_i}{R_1} \] - Multiply both sides by \( R_2 \): \[ v_o = \frac{R_2}{R_1} v_i \] - Therefore, \[ \frac{v_o}{v_i} = \frac{R_2}{R_1} \] --- **Final Answer:** \[ \frac{v_o}{v_i} = \frac{R_2}{R_1} \]